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Worked Examples · Example 7.15

Q.Although electron gain enthalpy of fluorine is less negative as compared to chlorine, fluorine is a stronger oxidising agent than chlorine. Why?

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Step 1 — Compare electron gain enthalpies alone.

ΔegH(F)=−333 kJ/mol,ΔegH(Cl)=−349 kJ/mol\Delta_{eg}H(F) = -333\ \text{kJ/mol}, \quad \Delta_{eg}H(Cl) = -349\ \text{kJ/mol}

By this factor alone, Cl should be the stronger oxidiser — but this ignores the other energy terms in aqueous oxidising ability.

Step 2 — Bond dissociation enthalpy.

F ⁣− ⁣F  bond enthalpy≈159 kJ/mol  (weak, due to small size + lone-pair repulsion)F\!-\!F \; \text{bond enthalpy} \approx 159\ \text{kJ/mol} \; (\text{weak, due to small size + lone-pair repulsion})

Cl ⁣− ⁣Cl  bond enthalpy≈242 kJ/molCl\!-\!Cl \; \text{bond enthalpy} \approx 242\ \text{kJ/mol}

Less energy is needed to atomise F2F_2 into atoms, favouring fluorine.

Step 3 — Hydration enthalpy.

ΔhydH(F−)  is much more negative than  ΔhydH(Cl−)\Delta_{hyd}H(F^-) \; \text{is much more negative than} \; \Delta_{hyd}H(Cl^-)

because the very small F−F^- ion is intensely hydrated, releasing far more energy on dissolution.

Step 4 — Net effect. …

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